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Old 2020-10-12, 05:33   #14
paulunderwood
 
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Sep 2002
Database er0rr

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Quote:
Originally Posted by Nick View Post
Feedback on version dated 11 October 2020 (wow you've written a lot!)
Thank you very much for taking tome to read the text. I have made some changes to my local copy but will upload after more changes.
Quote:
p5 The proof that root 2 is irrational relies on the uniqueness of prime factorization.
As this is a special property of the integers (not true in all number systems), I would at least mention it.
.
I mention it now in the both the paragraph on integers and when arguing about the irrationality of root 2.
Quote:
p6 Ordinary sets are unordered and may not have repeating elements (this is so that they correspond with properties - if you select all objects satisfying a certain condition, you want what you get to be a set). Orderings and multisets can be constructed from ordinary sets if needed (in fact, so can everything in mathematics).
I have added the word "not".
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p6 "subtracting all elements of one set from another" sounds confusing to me, as if you are calculating x-y for each x in the first set and y in the second. I would consider "removing"
I replaced the word "subtract" with "remove",
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p7 The group axioms \(e\circ g=g\) needs to be \(g\circ e=g\), or state it and the next one both ways around.
I have subtended e rather than prepending it and in all subsequent instances where identity existence in mentioned including the section on rings,
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p7 under multiplication the units of a commutative ring with 1 form an abelian group, but these are not all the elements of the ring (except in the trivial case where the ring has just 1 element).
I found this difficult. I have made mention of the center, although it might be out of context:
Quote:
\item multiplicative identity : $(\exists 1\in {\cal S})(\forall a\in{\cal S})\hspace{.1in}a\times 1=a$.
The units form an Abelian group apart from a ring with one element, called the {\em center}.
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p8 If you want to be able to relax condition 11 then you need to write condition 12 both ways around!
Done!
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p8 A field is also required to have 1 not equal to 0.
I have added "where $1\ne 0$".
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I'm out of time now - I'll look again in the week.
Again, many thanks,

Edit I have uploaded the latest copy -- 12 Oct.

Last fiddled with by paulunderwood on 2020-10-12 at 14:42
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